%I
%S 5,11,13,101,37,1277,1279,1616603,57405419,51448351,76623356077,
%T 115438255651991,433241801791933
%N Let p(k) denote the kth prime; a(n) = smallest p(m) > p(n) such that the n2 differences between [p(n), p(n+1), ..., p(2n2)] are the same as the n2 differences between [p(m), p(m+1), ..., p(m+n2)].
%e n=5: We take the four primes [p(5)=11, 13, 17, 19], whose successive differences are 2, 4, 2. The next time we see this sequence of differences is at [101, 103, 107, 109], so a(5) = 101.
%t (* This program generates the first ten terms of the sequence. To generate more would require significantly greater computing resources *) dbp[n_]:=Differences[ Prime[ Range[ n,2n2]]]; With[{prs=Prime[Range[ 3500000]]}, First/@ Flatten[ Table[Select[Partition[Drop[prs,n],n1,1], Differences[#]==dbp[n]&,1],{n,2,11}],1]] (* _Harvey P. Dale_, Feb 05 2014 *)
%o (PARI) A236411 = n>{d=vector(n2,i,prime(n+i)prime(n));forprime(p=prime(n+1),,for(k=1,#d,isprime(p+d[k])next(2));for(k=1,#d,p+d[k]==nextprime(p+if(k>1,d[k1])+1))next(2));return(p))} \\ The second kloop would suffice, but the first makes it 5x faster. Yields a(10), a(11) in ca. 3 sec (i7, 1.9Ghz).  _M. F. Hasler_, Feb 05 2014
%Y See A073615 for a very similar sequence.
%K nonn,more
%O 2,1
%A _Don Reble_, Feb 05 2014
%E Edited by _N. J. A. Sloane_, Feb 05 2014
